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Simple Stresses and
StrainsUNIT 1 SIMPLE STRESSES AND STRAINS
Structure
1.1 Introduction
Objectives
1.2
Basic Concepts
1.2.1 Rigid and Deformable Solids
1.2.2 Stresses and Strains
1.2.3 Tensile, Compressive and Shear Stresses
1.2.4 Complementary Shear Stresses
1.3 Mechanical Behaviour of Materials
1.3.1 Stress-Strain Curves
1.3.2 Hookes Law and Elastic Limit
1.3.3 Elastic Constants
1.4 Deformation of Bars
1.4.1 Bars of Uniform Section
1.4.2 Bars of Varying Cross Section
1.4.3 Bars of Uniform Strength
1.5 Composite Bars
1.5.1 Modular Ratio
1.5.2 Equivalent Area of a Composite Section
1.5.3 Stresses in Composite Bars and Load Carrying Capacity of Composite Bars
1.6 Temperature Stresses
1.6.1 Indeterminate Bars
1.6.2 Compound Bars
1.6.3 Yielding of Supports
1.7 Stresses on Oblique Sections
1.8 Relationship between Elastic Constants
1.8.1 Relationship betweenEand K
1.8.2 Relationship betweenEand G
1.8.3 Significance of the Relationships
1.9 Summary
1.10 Answers to SAQs
1.1 INTRODUCTION
You have already learnt how to analyse a given truss and find the forces induced in thevarious individual members. Later on you will be introduced to various other types of
structures and will learn how to find the forces induced in the members due to the forces
applied on the structure. Such analysis forms the essence of Structural Analysis. Our
concern in Strength of Material is what happens to the individual members, when they
have to carry these forces. As you advance in your learning of strength of materials, you
will realise that the two subjects are closely interrelated and together they form the
foundation for the design of a large variety of structures. In this unit, you will be
introduced to the analysis of simple stresses and strains induced in members.
Objectives
After studying this unit, you should be able to
understand the concepts of stress, strain and deformation,
elaborate the effect of forces on deformable solids and the relationshipbetween stress and strain for typical materials,
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Stresses in Solids describe the elastic properties of solids and their deformations due to directforces,
know the methods of analysing composite bars,
explain the deformations and stresses caused by temperature variation,
calculate stresses on oblique sections, and
define the elastic constants and their interrelationships.When you have clearly learnt these concepts, you will be able to use your knowledge to
understand the behaviour of solids under simple loadings and will be able to design such
members. You will also be able to verify whether such members are safe under the loads
they are subjected to. In addition, you will also be able to find the load carrying capacity
of members whose geometric and elastic properties are known.
1.2 BASIC CONCEPTS
1.2.1 Rigid and Deformable Solids
The subject matter of Strength of Materials is also learnt under the title Applied
Mechanics, and Mechanics of Solids. Applied Mechanics is the general title in which
Fluid Mechanics and Solid Mechanics are branches. What we study in Strength of
Materials is the effect of forces acting on solids of different geometrical and elastic
properties, and hence more people now choose to call it Solid Mechanics.
Let us now learn two terms used to describe two types of solids. A rigid solid is one
which does not undergo any change in its geometry, size or shape. On the other hand, a
deformable solid is one in which change in size, shape or both will occur when it is
subjected to a force. The geometrical changes produced are called deformations and
hence the name deformable solid. Many of the common solid articles we handle in our
day-to-day life exhibit perceptible deformations when subjected to loads. In many other
solids, though we may not perceive any deformations with our naked eye, with
measuring instruments of sufficient precision we can see that they also get deformedunder applied forces. A more careful scrutiny will reveal that all solids are deformable
and the idea of rigid solids is only a conceptual idealisation.
Look at the common swing shown in Figure 1.1, which you would have enjoyed playing
with during your earlier days. When a child sits on the swing we may not perceive any
appreciable deformations. But if the iron bars be replaced by bamboo sticks and the iron
chain be replaced by rubber strings, we can observe considerable deformations as shown
in Figure 1.2. We may then conclusively say that whenever forces are applied on solids,
deformations are introduced in them. All solids are deformable solids; hence, the subject
which once was called Mechanics of Deformable Solids is now simply called
Mechanics of Solids.
Figure 1.1
C D F
HT1 T2
W
E
G
E C D F
dc
e f
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Simple Stresses and
Strains
Figure 1.2
You may wonder as to why we should have the term rigid solid at all, if in reality, there
are no such solids. It has been already mentioned that the term rigid solid is only aconceptual idealisation. There are a few important uses for such idealization. Consider
the analysis of forces in the members of some of the plane trusses. You were given the
geometry of a structure (which in its totality may be considered as a solid of complex
geometry) on which some forces have been presented to act. While analyzing these
structures (solids) for finding out the member forces, the forces in the members have
been resolved into their horizontal and vertical components, assuming that the member
orientations have not changed due to the application of external forces. In reality, the
orientations of most of the members change, as the structure deforms under the loads.
However, these deformations and hence, the resulting changes in member orientations are
so small that the error in computed member forces are negligibly small. You may think
that even these errors may be eliminated by taking into account the deformations too. But
an attempt to do so will reveal the complexity of the calculations involved. Thecorrections thus effected are called secondary effects and may be neglected in most of
the cases. In the analysis of other types of structures also when the overall equilibrium is
considered the entire structure or any part under consideration may be treated as a rigid
solid.
In many cases, known as determinate problems, the analysis of forces within a solid can
be calculated treating it as rigid.
1.2.2 Stresses and Strains
When we apply forces on solids, deformations are produced if the solid is prevented from
motion (with acceleration = force/mass of solid) either fully or partially. If the solid is not
restrained, it may undergo displacements without change in shape or size and thesedisplacements are termed as rigid body displacements. If the solid is restrained by some
other force, known as reaction, which keeps the solid in equilibrium, the force will be
transmitted through the medium of the solid to the restraining support.
Consider the solid shown in Figure 1.3(a) which is pulled by the force P at the right hand
side and restrained by a support at the left hand side end. If the solid is to remain in static
equilibrium we know that the support should exert an equal pulling force in the direction
opposite to that of force P. This restraining force, termed as support reaction, is shown as
R in Figure 1.3(a). Now consider the solid being divided into a number of small bits or
elements of different lengths as shown in Figure 1.3(b). The force P applied at the RHS
end of element 1 is transmitted to the support through the medium of elements 2, 3 and 4.
The same thing is true even if you divide the solid into hundreds or even millions ofelements.
R P
Solid Bar
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Stresses in Solids
(a)
(b)
Figure 1.3
Now, to keep the element 1 in static equilibrium, element 2 should exert a force
R (= P) on element 1. At the same time, element 1 exerts a force P on element 2, so that
equilibrium exists at the point or surface of contact. The element 2 is again pulled by the
equilibrantR exerted by element 3. Thus equilibrium exists at each and every point or
surface of contact and each of the elements are kept in equilibrium.
The force P (orR) shown in the figures is only the resultant of forces applied over somearea (small or even full). On any area the force may not be distributed uniformly.
However, we may consider an infinitesimally small area A over which a small part ofthe total force P may be considered to be uniformly acting.
Here, the ratio0
or ratherA
dP PLt
dA A
may be called as pressure (suctional or
compressive) at the point. Considering a small element having the area dA on one of its
faces as shown in Figure 1.4, we realize that this pressure is accompanied by the
equilibrium pressure on the opposite face of the element.
Figure 1.4
Such a set of internal pressures acting on solids is defined as stress. Depending on which
set of planes such pressures are applied, there can be a number of such sets and they may
be called stress components (More about this we shall learn later).
To take the simplest case, if the total force P is uniformly distributed over the cross
sectionA as shown in Figure 1.3 (a), then the magnitude of stress in the solid is P/A. This
entity, stress is denoted by the symbol and usually expressed in N/mm units, also
denoted as Mega Pascal (MPa) as both are numerically equal.
Let us now turn our attention to deformations. The set of forces P andR
(Figure 1.3(a)) are pulling the solid in opposite directions. Consequently the solid getselongated and let this elongation be denoted by the symbol . If the forces be applied on
the individual elements, each element will elongate (Figure 1.5(b)) and it has been
experimentally established that
P
4 3 21
P P RR PRP
dp
dp
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Simple Stresses and
Strains(a) the magnitude of the individual elongations are proportional to the length of
each of the elements, and
(b) the total elongation of all the elements add up exactly to .
R P
L
(a)
(b)
PL1L2
L3L4
4 3 21
L
P
R (=P)
(c)
Figure 1.5
We may conclude that the ratio of elongation to length of the elements 1 2
1 2
, , . .L L
.
is
invariable and equal toL
. This ratio is defined as strain (more specifically as
longitudinal strain) and denoted by . If the forces applied are in the opposite direction,
as shown in Figure 1.5(a), the solid will get its length shortened and the deformation and
strain in such a case are considered negative.
1.2.3 Tensile, Compressive and Shear Stresses
We have observed that forces produce elongations or contractions on solids depending on
how they act. The forces that produce elongation on solids are called tensile forces and
the stresses induced by them are called tensile stresses. Tensile forces and tensile
stresses are both considered positive. Similarly, the forces which cause shortening of
length are called compressive forces and the stresses induced by them are called
compressive stresses. Both compressive forces and compressive stresses are considered
negative.
We may also observe that both tensile stresses and compressive stresses act normal to the
surface on which they act. For this reason, they are both classified as normal stresses.
However, there are many instances of load applications when the stresses are induced in
directions other than that of the normal to the surface. A few such cases are shown in
Figure 1.6. A small element is shown enlarged in scale. Besides the element, the
coordinate system is also given. The plane BEFC is normal to the x axis and is, hence,
defined asx plane. Similarly, the planes DCFG and ABCD are defined asy andz planes
respectively.
Y
FG
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Stresses in Solids
Figure 1.6
You may observe that the stress onx plane is normal (tensile), stress ony plane is
inclined and that onz plane is parallel to the plane itself. The stresses that are acting
parallel to the plane on which they are applied are called shear stresses. On further
consideration we may resolve the stress on they plane into components parallel and
normal to the surface. Hence, on any plane, there may be normal stresses, shear stresses
or both. Further, while the direction of normal to a plane is uniquely defined, there areinfinitely large numbers of directions in which shear stresses may be applied on a plane.
Whatever may be the direction of shear stress, it can be further resolved into components
in two mutually perpendicular directions, and having done this, you would have
completely defined the stresses acting on a plane. Defining such states of stress on all the
planes will furnish the complete state of stress on the small element (or at a point).
Let us now consider a few examples of solids in which shear stresses are induced.
Figure 1.7 shows how two steel plates are connected together by a riveted joint. When
the connected plates are pulled, the pull from plate A is transmitted to plate B through the
rivet. The rivet may be considered to consist of two parts. Part C of the rivet carries the
force P from plate A by diametric compression or bearing, while part D of the rivettransmits the load P to plate B by bearing. The transfer of force from part C to its
counterpart D takes place at the common interface namely its cross sectional area at
middle length. On this area, the force P is applied parallel and hence, the rivet is in shear.
Again consider two wooden blocks connected together by some glue and also connected
to a rigid base similarly, as shown in Figure 1.8. If a horizontal force P is applied on thex
plane of the block A at the top, this force is transmitted to the base through a horizontal
surface which is the common interface between the base and the block D. As the force is
parallel to the surface on which it is acting, it is shear force and the stress induced by it
also should be shear stress. On further scrutiny, you will find that the transmission of
force from one block to another in the assembly is only as shear force and not only on
interfaces between two adjacent blocks, but on any horizontal.
Figure 1.7(a)
P
AP
B
P P
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Simple Stresses and
Strains
(b)
(c)
P
P
(d)
Figure 1.7
Figure 1.8
a b
dc
h
e P
X
f
g
Qi
l
po
nR
k
j
m
X
X
X
X = X
A b c d etc. are Cornersof Different Blocks
1.2.4 Complementary Shear Stresses
Consider the equilibrium of the solid element of dimensions l, b and h and acted upon by
shear stress components xy and yx as shown in Figure 1.9. By inspection, we may
recognize that the components xyon a pair of vertical planes and the pair ofyxon two
horizontal planes satisfy independently the force equilibrium equations :
Fx = 0 and Fy = 0
Let us consider the moment equilibrium equations. xycomponents on planes CDHG and
ABFE have resultant forces Qxy = xy . b. h which form a couple equal to (xy . b . h) l.
Similarly, the moment of the couple formed by the shear stress component
xy = (yx. l . b) h. For equilibrium,
(xy. b . h) l + (yx . l . b) h = 0
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or xy = yx.Stresses in Solids
That is xyis always accompanied by yxand this pair of shear stresses are called
complementary shear stresses.
l
Figure 1.9
When a piece of carrot is cut into two pieces, the force exerted by the knife acts on
surface parallel to the force and hence the carrot is sheared into two. Likewise, you may
think over the various cases of load applications in our day-to-day life and identify the
cases in which shear stresses are involved.
1.3 MECHANICAL BEHAVIOUR OF MATERIALS
1.3.1 Stress-Strain Curves
As we have earlier observed, the behaviour of solids, when subjected to loads, depends
upon its geometrical and mechanical properties. The mechanical properties of solids have
to be investigated through laboratory tests. Let us consider one such test so that you can
understand certain mechanical properties of solids.
Tension Test on a Mild Steel Rod
A mild steel rod of known diameter and length of about 400 mm to 500 mm is
firmly gripped in a universal testing machine and a deformeter is fixed on the rod
over a measured gauge of 150 mm to 200 mm. The rod is subjected to axial tensile
load. The load at any instant is indicated by the load dial of the UTM while theelongation of the bar over the gauge length is measured through the deformeter.
The load is gradually applied on the bar in suitable increments and the
corresponding elongations of the bar are measured at each stage. The test is carried
out until the specimen ultimately fails by rupture. The load values at each stage are
divided by the cross sectional area of the bar to get the stress values and the
corresponding elongations of the bar divided by the gauge length give the strain
values. A graph is drawn relating stress and corresponding strain. A typical stress-
strain curve for mild steel is shown in Figure 1.10.
b
D
H
B C
E
FG
A
yx
yx
xy
xy
Y
Z
h
X
Y
Stress
Cardinal Points
A : Proportionality Limit
B : Elastic Limit
C : Yield Point
C D : Ductile Extension Zone
E : Ultimate Stength
F : Ru ture Stren th
A
B C
D F
E
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effecting permanent changes in the geometry of the solid. The strain
remaining unrelieved is known as permanent set.
Stresses in Solids
(c) When we reload a solid which is already subjected to permanent set, thestress-strain curve for reloading is linear up to the maximum stage to which
it was subjected to earlier.
From these observations, what does a designer learn? If a structure or any component is
loaded beyond proportional limit, the geometry gets permanently changed and if
repeatedly loaded beyond proportional limit, it will go completely out of shape after a
few repetitions. On the other hand, you can load it within their proportional limit any
number of times and the original geometry recover on the removal of the load. Hence, in
any design, one has to be careful to avoid loading the component beyond the proportional
limit. Within this limit the solid is said to be elastic. The property called elasticity
consists of two aspects, normally complete removal of strains on unloading and linearity
of stress-strain relationship. Hence, the part of stress-strain curve up to yield point is also
called as elastic limit. To ensure that the solid is always within this elastic limit,
designers permit stresses in solid much lower than the elastic limit. The ratio of the
maximum stress a material can withstand to the maximum stress which a designer
chooses to allow is called the factor of safety. In short,
Factor of Safety =Maximum stress the material can withstand
Maximum stress permissible
If the numerator is yield point stress, a factor of safety is 2 usually allowed and if it is
ultimate stress, factor of safety may be 3 or more.
All materials do not behave as mild steel. The stress-strain curves of a few materials are
given in Figure 1.12. Irrespective of the variety of shapes seen in these curves, one can
identify in each of them a segment where the curve may be considered linear. This limit
is taken as the elastic limit for the material.
200
400
600
800
0.1 0.2 0.3 00 .4 0.5
Manganese Steel
Nickel Steel
Bronze
Aluminium
Stress
(N/mm
2)
Strain
Figure 1.12 : Stress-Strain Relationship for a Few Materials
The overall behaviour of the material and its complete stress-strain relationship is
discussed here only in order to caution the designer to keep the stress level well within
the elastic limit. Hence, for practical purposes the stress-strain relationship is taken as
linear and during the rest of this course you will not be dealing with nonlinearity of
material behaviour. Hence, for a designer, stress is always proportional to the strain andthe proportionality constant is defined as Elastic Modulus.
Thus, Stress, = Elastic Modulus (E) Strain ()
or =E. . . . (1.1)
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Simple Stresses and
StrainsFrom Eq. (1.1), we get
E=
. . . (1.2)
Eq. (1.2) indicates that the Elastic Modulus of a material may be calculated by measuring
the slope of the stress-strain curve (within the elastic limit). The fact that
Eq. (1.1) holds good within elastic limit which has already been taught to you as
Hookes Law, which we may restate as follows :
Within elastic limit the strain produced in a solid is always proportional to the stress
applied.
1.3.3 Elastic Constants
As there are different types of stresses possible, such as linear (normal) stress, shear
stress, volumetric (bulk) stress, and correspondingly different modulii of elasticity
namely Youngs modulus, shear modulus and bulk modulus are defined as elastic
constants of a material. The termEis used in Eqs. (1.1) and (1.2) in the context of normal
stresses and corresponding strains is known as Youngs Modulus.
In addition to the three modulii of elasticity an important elastic constant, used indefining the mechanical properties of a solid, is known as Poissons ratio. While testing
the material for stress-strain relationship, you observe that strains are produced not only
in the direction of the applied stress, but also in direction perpendicular (lateral) to it. On
further investigations, you will find that the lateral strain is always proportional to the
longitudinal strain and the proportionality constant is negative. This proportionality
constant is defined as Poissons Ratio and denoted by .
Poissons Ratio, =StrainalLongitudin
StrainLateral
While applying tensile force on a rod we found that the deformations produced involved
change of volume of the solid as well as its shape. Such deformations may be found in
many other cases of loading. Deformations involving change of volume and shape are
more common. However, we may identify two special cases of deformations, dilatation
and distortion.
Dilatation is defined as change in volume (bulk) of the solid without change of shape. If
normal stress components of equal magnitude and same nature (all tensile or all
compressive) are applied on all the three mutually perpendicular planes of a solid
element as shown in Figure 1.13, the solid will undergo only volume change without any
change in shape. Such a state of stress (as may be obtained in the case of solid subjected
to hydrostatic pressure) is called volumetric stress and the change of volume produced
per unit volume of the solid is called the volumetric strain. The ratio between
volumetric stress and volumetric strain is defined as Bulk Modulus.
Thus, Bulk Modulus, K=StrainVolumetric
StressVolumetric
. . . (1.3)
in which Volumetric strain =VolumeOriginal
VolumeofChange
The state of stress, with equal values of stress components in all the direction is also
known as spherical state of stress.
y = O
BxB = BOB
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Stresses in Solids
Figure 1.13 : Bulk Stress and Dilatation
The state of deformation involving change of shape without any change of volume is
defined as Distortion. Though distortion may be introduced in many ways let us consider
a simple case. When shear stresses are applied on a solid as shown in
Figure 1.14, we find that angular deformations are introduced. The angular strain or
change in angle produced is called shear strain denoted by (Greek gamma) and itsmagnitude is expressed in radians. The ratio of the applied shear stress and the shear
strain produced by it is defined as Rigidity Modulus or Shear Modulus of the material,
i.e.
Rigidity Modulus, G =)(StrainShear
)(StressShear
. . . (1.4)
Figure 1.14 : Shear Stress and Shear Strain
We have so far defined four elastic constants, namely, Youngs Modulus (E), Poissons
Ratio (), Bulk Modulus (K), and Shear Modulus (G). Later on we shall learn that these
four are not independent constants, but are related to each other. We can show that there
are only two independent elastic constants and the other two are dependent on them.
1.4 DEFORMATION OF BARS
1.4.1 Bars of Uniform SectionWith the knowledge gained already, you will now be able to learn calculation of
deformations in solids due to applications of simple stresses. Let us start with the
example of a prismatic bar of length L, uniform cross section of areaA carrying an axial
load P as shown in Figure 1.15.
L
P
P
Area ofCross-section A
Figure 1.15
Stress in the solid is in the longitudinal direction and its magnitude may be calculated as
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Simple Stresses and
StrainsA
P==
Area
Load
If the Youngs Modulus of the material,E, is known, the strain induced may be
calculated as
AE
P
E=
=
As strain is only deformation per unit length, the total elongation of the bar is calculated
as
= .L = LAE
P
Thus, Elongation =AE
PL. . . (1.5)
You may please note that the strain, , calculated here is in the longitudinal direction and
is accompanied by strains in the lateral directions also whose magnitude is given by .Let us now consider a few numerical examples.
Example 1.1
If the bar shown in Figure 1.15 is 1 m long with rectangular cross section of
300 mm deep and 400 mm wide, calculate the change in volume of the solid due to
a longitudinal compressive force of 720 kN, if the elastic constantsEand for the
material are known as 120 kN/mm2
and 0.2 respectively.
Solution
Area of cross section of the member = 300 400 = 120000 mm
Longitudinal strain =AE
P=
310120120000
1000720
= 0.00005
(Note that all the values have to be converted to consistent units; here, it is N for
forces and mm for length.)
Total change in length = 1000 ( 0.00005) = 0.05 mm.
Lateral strain l= = 0.2 ( 0.00005) = 0.00001
Change in depth = 0.00001 300 = 0.003 mm
Change in width = 0.00001 400 = 0.004 mm
Change in volume of the solid,
= (1000 0.05) (300 + 0.003) (400 + 0.004) (1000 400 300)
= 999.95 300.003 400.004 (1000 400 300)
= 3600.108 mm
Let us consider an alternate approximate method also.
Change in volume, dV= (V+ dV) V
= (l + l) (b + b) (d+ d) l . b . d
where l, b, and dare changes in length, breadth and depth of the solid.
i.e. dV= l (1 + 1) b (1 + 2) d(1 + 3) l . b . d
where 1, 2 and 3 are the strains in the three mutually perpendicular directions.
dV= l bd (1 + 1) (1 + 2) (1 + 3) l bd
= l bd (1 + 1 + 2 + 3 + 12 + 23 + 31 + 123) l bd
= l bd (1 + 2 + 3 + 12 + 23 + 31 + 123)
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Neglecting the second order products,Stresses in Solids
dV= V (1 + 2 + 3) . . . (1.6)
Now let us calculate the change in volume of the given solid using Eq. (1.6).
Change in volume, dV= V (1 + 2 + 3)
= 1000 300 400 ( 0.00005 + 0.00001 + 0.00001)
= 3600 mmThough there is a small error, the approximation is quite satisfactory (As an exercise you
may calculate the percentage error in the value). If you are very particular about
accuracy, you use the following formulation:
dV= V (1 + 2 + 3 + 12 + 23 + 31 + 123)
SAQ 1
(a) Taking the dimensions of the solid in Example 1.1 as same, calculate thechange in volume of the solid if additional tensile forces of
1200 kN, and 2000 kN are applied in the lateral directions namely horizontal
and vertical respectively.
(b) A concrete cylinder of height 300 mm and diameter 150 mm is tested forcompression in a Universal Testing Machine. Within the elastic limit, the
cylinder was found to be shortened by 0.12 mm and its diameter was found
to be increased by 0.01 mm under an axial load of 90 kN. Calculate the
Youngs Modulus,E, and Poissons Ratio, , for the specimen.
1.4.2 Bars of Varying Cross SectionLet us now consider a few cases of a little more complexity. To give you a feeling of
solid reality all solids have so far been represented by pictorial views. Figure 1.16
represents the projected view of a bar whose cross sectional area differs in different
segments. Taking the elastic modulus of the material as 60 kN/mm, let us evaluate the
total elongation of the bar.
200 135 60 100750 kN
300 mm
750 kN
400 mm 200 mm 500 mm
(Numbers in Members Indicate Diameters in mm)
Figure 1.16
Although the bar has no uniform cross-section, it consists of segments of uniform cross
section. Hence, the total elongation of the bar can be calculated as a sum of the
elongations of individual segments.
Thus, = i=1 1 2 2
1 1 2 2
... ...i i
i i
PLP L P L
A E A E A E+ + + + +
In the numerical example under consideration
=
601004
500750
60604
200750
601354
400750
602004
300750
2222
+
+
+
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Simple Stresses and
Strains=
+++
2222 100
500
60
200
135
400
200
300
604
750= 2.14865 mm
In the bar, shown in Figure 1.16, the axial pull is applied at the ends and hence the axial
force in all the members is same. On the bar, shown in Figure 1.17(a), the external forces
are applied at intermediate sections also. In such cases, the axial force in each member
should be evaluated first, before any deformation calculations. This can be accomplishedby considering equilibrium of each of the sections where external loads are applied.
Though no external load has been prescribed at the LHS end, the support reaction has to
be calculated and taken as the external load. Equilibrium analysis can be easily carried
out by treating each segment as a free body as shown in Figure 1.17(b).
For example, consider the equilibrium of the segment 4 in Figure 1.17(b). At the RHS
end of the member a point load of 60 kN is applied. Hence, for the member to be in
equilibrium a force of 60 kN should be applied at the RHS end of the member. Hence,
the member is subjected to a tensile force of 60 kN which is represented by the internal
arrows in accordance with the sign conventions you have already learnt. Member 3 is
pulled with a tensile force of 60 kN exerted by member 4 and in addition the external
force of 80 kN also pulls the member in the same direction, resulting in the member
carrying a total tensile force of 140 kN. Proceeding thus, the axial forces in all the
members can be calculated. To simplify the graphical representation we may show the
member forces along with external forces as shown in Figure 1.17(c).
Now let us calculate the total elongation of the bar, taking the elastic modulus,E, as
200 kN/mm.
L = 800 mm
A = 200 mm
2L = 600 mm
A = 800 mm
2L = 800 mm
A = 600 mm2
L = 500 mmA = 150 mm
2
60 kN 40 kN 80 kN 60 kN
(a) Not to the Scale60 kN
80 kN40 kN60 kN
(b)
1 2 3 4
(c)
110 kN
160 kN 100 kN 140 kN 60 kN
60 kN80 kN40 kN60 kN
Figure 1.17
= i = ii
ii
EA
LP
=200150
50060
200600
800140
200200
800100
200800
600160
+
+
+
= 0.6 + 2.0 + 0.9333 + 1.0
= 4.53333 mm.
Now, let us consider the case of a bar whose cross section is varying continuously as
shown in Figure 1.18. The bar of (truncated) conical shape is subjected to an axial force
ofP. As in the case of stepped bars, here also, we may apply the equation
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20
= i = ii
ii
EA
LP, if we recognize thatLi is sufficiently small andAi may be taken as
uniform over the length of the small segment. That is, the expression ii
ii
EA
LPnow
becomesii
ii
L
EA
LP0
.
Stresses in Solids
P
L
PD2D1
Figure 1.18
Figure 1.19
However, to simplify the expression forA, the origin may be conveniently be shifted to
the apex of the produced cone as shown in Figure 1.19. Thus,
0
21
4
H
H
Pdx
D xE
H
=
The integral in above equation may be evaluated for solids having stepped as well as
continuous variation in cross section. Stepped bars require no special treatment except
that the equilibrium and deformations of each segment may be analysed separately and
added wherever necessary. Even a prismatic segment of the bar may have to beconsidered as two or more members, if external loads are applied at interior points of the
segment.
SAQ 2
(a) Taking the Elastic modulus,E= 80 kN/mm and Poissons ratio of the bar as
0.24, calculate the change in volume of the bar shown in Figure 1.16.
(b) Calculate the total elongation (or shortening) of the non-uniform bars (with
loads) shown in Figures 1.20 and 1.21.
D1H =D -D1 2
L
X
dx
D1
Ho
D2
d = 10 mmd = 8 mm d = 5 mm
600 mm 1200 mm 800 mm
440 N
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Figure 1.20 : (E = 180 GPa)
Simple Stresses and
Strains
(360)4800 N
(300)
Figure 1.21 : (E = 200 GPa)
7200 N 3000 N 3600 N(480)
(240)
180 150 160 120
A B C D E
(c) Taking the Youngs Modulus,Eas 120 kN/mm, find the total elongation of
the bar shown in Figure 1.22.
80 kN 30 kN 50 kN120 kN
d = 28 mmd = 50 mm
600 mm 800 mm 700 mm 700 mmd = 60 mm
Figure 1.22
1.4.3 Bars of Uniform Strength
If we calculate the stresses induced in the various segments of the bar shown in
Figure 1.17(a), we get the following results :
1 =160 1000
800
= 200 N/mm
2 =100 1000
200
= 500 N/mm
3
=140 1000
600
= 233.33 N/mm
4 =60 1000
150
= 400 N/mm
The results show that the bar should be made of a material with permissible stress not
less than 500 N/mm. Such a material (should be specially manufactured, if required) is
wasted in segments where the stress induced is much less. For instance, in segment 1, the
stress induced is only 200 N/mm. If we reduce the cross sectional area of segment 1, 3
and 4 to 320 mm, 280 mm and 120 mm respectively, the stress in all segments will be
uniformly 500 N/mm2, the bar would be safe and we would have effected considerably
economy of material. Such a bar is called a bar of uniform strength. A designer will strive
to achieve a bar of uniform strength, whenever possible. However, due to otherconstraints and variations in loadings, it may not always be possible to provide a bar of
uniform strength. Thus, the concept of bar of uniform strength is a design ideal aiming to
effect maximum saving in the material.
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Stresses in Solids1.5 COMPOSITE BARS
So far we have considered the cases of solids made of a single material. However, in a
number of engineering applications we use components made of two or more materials.
Bars made of two or more materials are called composite bars. This section will
introduce to you basic principles applied in analyzing such bars.
1.5.1 Modular Ratio
Consider the composite bar shown in Figure 1.23 in which an aluminum bar is encased
just snug in a copper tube which in itself is encased just snug in a steel tube. Let us
analyze how the three different components of the composite bar share the load applied
axially on the composite bar.
Aluminum
Copper
Steel
50 mm 90 mm 120 mm60 kN
400 mm
Figure 1.23
The basic requirement in such a case is that all the three components of the bar shouldundergo the same amount of deformation while carrying the load. Otherwise, the bar
which shortens the least alone will be in contact with the load which is physically
impossible, since a component which looses its contact will not be carrying any load and
hence there can be no shortening of that component which contradicts the assumption
that it shortens more. Hence, it is logically necessary that all the three bars should
undergo the same amount of deformation. This condition is called the compatibility
condition or condition of consistent deformation. Satisfaction of this condition is the
basis on which the problem of composite bars is analysed.
Let the total axial force applied on the composite memberP be shared by the different
members as P1, P2 and P3. If the deformations produced are 1, 2 and 3,then we get,
1 =11
11
EA
LP
2 =22
22
EA
LP
3 =33
33
EA
LP
The compatibility condition requires that
11
11EA
LP=
22
22EA
LP=
33
33EA
LP. . . (1.7)
Let the elastic modulii of the materials be redefined asE1, m2E1 and m3E1,where
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23
Simple Stresses and
Strainsm2 =1
2
E
Eand m3 =
1
3
E
E
Eq. (1.7) may be rewritten as follows
3
3
13
3
2
2
12
2
1
1
1
1
A
P
Em
L
A
P
Em
L
A
P
E
L== . . . (1.8)
Recognising thatL1 =L2 =L3, we may rewrite Eq. (1.8) as
33
3
22
2
1
1
Am
P
Am
P
A
P== . . . (1.9)
from which, we get P2 =1
1
A
Pm2A2 and
P3 =1
1
A
Pm3A3 . . . (1.10)
The equilibrium condition required to be satisfied is
P1 + P2 + P3 = P
or P1 +1
1
A
Pm2A2 +
1
1
A
Pm3A3 = P
or P1
++
1
33
1
221A
Am
A
Am= P
P1 =
++
1
33
1
221A
Am
A
Am
P. . . (1.11)
After calculating P1 using Eq. (1.11), the loads shared by other members P2 and P3 may
be calculated using the Eq. (1.10).
In the process of the above solution, we have introduced two terms, namely, m2 and m3
whose values are the ratios1
2
E
Eand
1
3
E
Erespectively. These ratios are the ratios of
Elastic modulii of two different materials and, hence, are called Modular Ratios.
Let us now solve the problem of load sharing in the composite member shown in
Figure 1.23.
Here, we have
P = 60 kN
L = 400 mm
A1 =4
50
2= 1963.5 mm
2
A2 =4
(90
2 502) = 4398.23 mm2
A3 =4
(120
2 902) = 4948 mm2
Elastic moduli for the materials, namely aluminium, copper and steel may be taken as 80
kN/mm, 120 kN/mm and 200 kN/mm respectively.
m1=80
120
1
2 =E
E= 1.5
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Stresses in Solidsm2 =
80
200
1
3 =E
E= 2.5
Substituting these values in Eq. (1.12),
P1 =
++
5.1963
49485.2
5.1963
23.43985.11
60
= 5.6285 kN
Now, using Eq. (1.10),
P2 =5.1963
62852.5 1.5 4398.23 = 18.912 kN
P3 =5.1963
62852.5 2.5 4948 = 35.4595 kN
1.5.2 Equivalent Area of a Composite Section
On reconsideration, we may rewrite Eq. (1.11) as follows :
P1...][ 3322
1
1 +++=
AmAmA
PA
or...][ 33221
1
1 +++=
AmAmA
P
A
P. . . (1.12)
In Eq. (1.12), we recognize that1
1
A
Pis the stress induced in material 1 and hence,
1...][
33221 +++
=AmAmA
P. . . (1.13)
The denominator in Eq. (1.13) has Area unit and is called the equivalent area of the
composite bar, as if composite bar entirely made of material 1 alone. In other words, if
the composite bar in Figure 1.23 is entirely made of aluminum, then the equivalent area
is given as
Aeq =A1 + m2A2 + m3A3
= 1963.5 + (1.5 4398.23) + (2.5 4948)
= 20930.86 mm
That is, the composite bar, shown in Figure 1.23, is equivalent to an aluminium bar of
cross sectional area 20930.86 mm, in resisting axial forces. (An aluminium bar of 163.25
mm diameter will give this area.)One of the practical examples of composite bars is the RCC column, a typical cross
section of which is shown in Figure 1.24. Here, a 400 mm 400 mm square cross section
of a RCC column is reinforced with 8 numbers of 32 mm diameter mild steel bars.
Taking the Elastic modulii of steel and concrete asEs andEc and the modular ratio, m =
C
S
E
Eas 18, let us find the loads shared by steel and concrete when an axial load of 600 kN
is applied on the column.
Area of steel bars,As = 8 4
32 = 6434 mm
Area of concrete, Ac= (400 400 6434) = 153566 mm
Modular Ratio, m= 18
Equivalent Area of Concrete = (153566 + 18 6434) = 269378 mm
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Simple Stresses and
StrainsTotal axial load = 600 kN.
Load shared by concrete =269378
600153566 = 342.046 kN
Load shared by steel =269378
600 18 6434 = 257.954 kN
RCC Column Cross Section
(400 mm 400 mm with 8 No. of 32 mm MS Bars)
Figure 1.24
SAQ 3
(a) Show that the composite bar in Figure 1.23 is equivalent to a steel bar ofcross sectional area 8372.344 mm.
(b) If the composite bar shown in Figure 1.23 is to be replaced by a copper bar,calculate the equivalent area of cross section of the copper bar.
1.5.3 Stresses in Composite Bars and Load Carrying Capacity of
Composite Bars
You have learnt how a total load applied on a composite bar is shared by different
components of a composite. Though this is certainly enlightenment on the behaviour of
composite bars, we need to learn more in order to apply the concept in practically more
useful way. That is, we need to learn
(a) what will be the total load that a composite bar can carry, and
(b) to carry a given load, how should a composite bar be proportioned.
For this purpose, we should know the strength of each material and also we should be
able to calculate the stresses induced in the various components of a composite bar.
We have already learnt the compatibility condition that the axial deformations undergone
by all the components of a composite bar should be equal.
As the lengths of components are also equal, the strain in each component should also be
equal. Taking the strain of the composite bar as , stress in any component may be
expressed as
i=Ei . . . (1.14)
or i=1E
Ei E1
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or = mi E1 Stresses in Solids
Since, E1 = 1, then we can write
i= mi 1 . . . (1.15)
Eq. (1.15) expresses the relationship that the stress induced in any component of a
composite bar should be proportional to its elastic modulus or modular ratio. For
example, in the RCC column shown in Figure 1.24, the stress in steel, s,will be
18 times the stress in concrete. You could have seen this by yourself, if you had dividedthe load shared by each member by its area of cross section. Use of Eq. (1.15) saves some
computational effort.
We can get stress in concrete =153566
1000046.342 = 2.227355 N/mm.
Thus, Stress in steel = 18 2.227355
= 40.0924 N/mm.
In order to calculate the load carrying capacity, we need to know the strength (allowable
stress) of all the components. For example, let the allowable stress in concrete be
4 N/mm and the allowable stress in steel be 120 N/mm. There is another restriction, youcan notice, that even though the strength of steel is 120 N/mm, we cannot stress it to the
full, because if we stress steel to its full strength, i.e. 120 N/mm, then concrete will be
stressed to 6.667 N/mm, i.e. 120/18, which is not permissible. Hence, the maximum
stress that may be induced in steel is only 72 N/mm, i.e. 4 18.
Now, we can calculate the maximum load the column can support or the load carrying
capacity of the column as follows :
P = s As + c Ac
= 72 6434 + 4 153566
= 1077512 N or 1077.512 kN.
Another type of problem that a designer faces is to decide the amount of steel required if
the column is to support a magnitude of load. For instance, if the column is required to
carry an axial load of 960 kN, let us calculate the steel requirement.
Maximum stress in steel = 72 N/mm
Maximum stress in concrete = 4 N/mm
Let the area of steel beAs.
Area of concrete = 160000 As.
72 As+ 4 (160000 As) = 960 1000(72 4)As+ 640000 = 960000
As=( )
472
640000960000
= 4706 mm
This may be provided suitably (say, 8 nos of 28 mm dia bars, which will be a little more
than sufficient).
SAQ 4
(a) Evaluate the stress induced in each component of the composite bar shownin Figure 1.23.
(b) In the case of composite bar shown in Figure 1.23, if the permissible stressesin aluminium, copper and steel respectively are 50 N/mm
2,
60 N/mm2
and 120 N/mm2, calculate the load carrying capacity of the
composite bar.
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Simple Stresses and
Strains(c) If the column shown in Figure 1.24 is required to carry an axial load of
880 kN, find the requirement of steel, with mechanical properties as given in
worked example.
(d) Provide a graphical chart to relate the strength of column shown in
Figure 1.24 to the percentage ratio ofA
AS whereA =AS +AC. Take
C= 4 MPa, S = 140 MPa and m = 18.
(e) If the column shown in Figure 1.24 is cast with higher quality of concretewith a permissible stress of 5 N/mm
2andEC= 0.06ES, what is its load
carrying capacity and what is the quantity of steel required if the column has
to carry an axial load of 1200 kN?
1.6 TEMPERATURE STRESSES
You have studied in physics about the expansion of solids due to rise in temperature. You
may recall that the linear expansion of a solid is proportional to the rise in temperature
and the proportionality constant is a material property called Coefficient of Linear
Thermal Expansion denoted by , its units are expressed as m/m/oC. If is thecoefficient of linear thermal expansion of a solid, its linear dimensions will increase by m per every meter of its original length for every 1
oC rise in temperature.
We may symbolically represent this as,
. . . (1.16)L T =
where, is the rise in temperature inT oC units.
When the temperature of a solid is raised, expansion or thermal deformation is certainly
produced, but it need not always produces stresses.
Thermal stresses are produced in a solid only if the expansion is restrained either fully or
partially. This section will introduce to you the methods of analysis of thermal stresses in
solids.
1.6.1 Indeterminate Bars
With reference to thermal stresses, bars whose thermal expansion is not restrained in any
manner are called determinate bars and the bars whose thermal expansion is restrained
in any manner are called indeterminate bars. As in the case of many of the other
indeterminate problems thermal stress analysis of indeterminate bars may be carried out
easily by the method of compatibility or consistent deformations.
Consider the prismatic bar of length L shown in Figure 1.25. The bar is connected to a
rigid support at LHS and free at the other end. But at a small distance d from the free
end of the bar there is a rigid body which would prevent any expansion of the bar beyond
the small gap of which d. If the thermal expansion of the bar is less than or even equal
to d no stresses will be induced in the bar and the problem remains determinate. If the
bar is heated further, then temperature stresses will be produced. As the bar is heated it
will push against the rigid support which will restrain the expansion by exerting a
compressive force on the bar, which must produce elastic shortening equal to the excess
thermal expansion.
L d
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28
Stresses in Solids
Figure 1.25
Let us illustrate the principle by a numerical example. Let the length and cross-sectional
area of the bar be 750 mm and 1200 mm2
respectively. Let us also assume, d= 0.12 mm,
E= 200 kN/mm2, = 12 10 6 m/m/oC and T = 32oC. Using the above data, we canreadily calculate the free thermal expansion of the bar is
t L T =
6750 12 10 32 0.288 mm= =
But this free expansion cannot take place due to the presence of the rigid restraint after a
gap of 0.12 mm.
Assuming the elastic force in the bar as P, the elastic deformation e may be calculated as
e
PL
AE =
Any how the total deformation of the bar should be only equal to d.
Hence, the compatibility condition may be expressed as
t e d + = . . . (1.17)
( )PL
L TAE
d + =
From which ( )AE
P L T d L
= . . . (1.18)
ord
P TL
=
AE . . . (1.19)
Substituting the numerical values in Eq. (1.18),
1200 200(0.288 0.12)
750P
=
= 53.76 kN.
Now, stress induced253.76 1000 44.8 N/mm
1200t
= =
In many cases, the bar may already be kept between supports and d is zero so that wemay write as follows :
P T AE =
or thermal stress t E T = . . . (1.20)
Eq. (1.20) implies that the thermal stress induced is a function of rise in temperature and
elastic and thermal properties and not dependent on the geometric properties of the bar.
SAQ 5
(a) If the non-uniform bar shown in Figure 1.16 is placed between two rigidsupports with a small gap of 0.2 mm and the bar is heated by 30
oC, calculate
the maximum thermal stress induced in the bar.Take = 12 10 6 m/m/oC andE= 2000 GPa.
(b) If the non-uniform bar shown in Figure 1.17 is placed just snug between tworigid restraints, calculate the thermal stresses induced in the bar due to
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29
Simple Stresses and
Strainstemperature rise of 24
oC. andEmay be taken as 12 10 6 m/mo/C and
200 GPa.
(c) If the permissible stress in the material of the bar shown in Figure 1.17 is105 N/mm
2, find out the maximum temperature rise to which the bar may be
safely subjected to.
1.6.2 Compound Bars
In the case of thermal stress analysis in compound bars, the following points should be
clearly understood and carefully applied :
(a) When the temperature of the compound bar is raised, each of its componentstend to expand differently, depending on their values.
(b) How much to the thermal expansion in each component is decided based oncompatibility conditions.
(c) When once the deformation to be suppressed in the case of each componentis known, the stress induced may be calculated for each bar according to its
Evalue.
Example 1.2
Consider the compound bar shown in Figure 1.26 consisting of a steel bolt of
diameter 18 mm, surrounded by a copper tube of outer and inner diameters 30 mm
and 18 mm respectively. The assembly is just snug at 15oC. The materialproperties are given as :
Youngs modulus of steel,ES = 200 kN/mm2
Youngs modulus of copper,EC= 120 kN/mm2.
Coefficient of linear thermal expansion of steel, 6 o12 10 m/m/ Cs =
Coefficient of linear thermal expansion of copper, 6 o18 10 m/m/ Cc =
Calculate the thermal stresses in steel and copper when the temperature of the
assembly is raised to 45oC.
160 mm
BoltHeart
Copper TubeSteel Bolt
Nut
Figure 1.26
Solution
In this case, when the temperature of the assembly is raised, copper will expand
more than steel and hence, will exert a thrust on the steel nut and bolt head thus
producing tension in the steel. Since there is no other external force this tension
has to be kept in equilibrium by the compressive force induced in copper. Hence
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30
by formulating the compatibility condition of total strains (thermal + elastic) the
problem can be solved.
Stresses in Solids
Let the forces induced in steel and copper be designated as Ps and Pc respectively
in kN units.
Thermal strain in steel, st s T =
612 10 (40 15)=
43 10 m/m=
Thermal strain in copper, c T
618 10 (40 15)=
44.5 10 m/m=
Elastic strain in steel, sses s
P
A E =
218 2004
sP=
51.965 10 sP=
Elastic strain in copper, ccec c
P
A E =
2 2(30 18 ) 120
4
cP=
51.8421 10 cP=
Compatibility Condition
Total strain in steel = Total strain in copper
i.e. 4 5 43 10 1.965 10 4.5 10 1.8421 105
s cP P + = +
Recognising that Pc = Ps, and dividing both sides by 10 5
, we get
30 + 1.965 Ps = 45 1.8421 Ps
or
45 30
3.94 kN(1.965 1.8421)sP
= =+
Stress in steel 22
3.94 100015.5 N/mm
184
= =
Stress in copper 22 2
3.94 10008.71 N/mm
(30 18 )4
= =
SAQ 6
(a) The compound bar shown in Figure 1.23 is placed between two rigidsupports with a small gap of 0.1 mm and its temperature is raised by 24 oC.Calculate the thermal stress in Aluminium, Copper and Steel if
6 o23 10 m/m/ Ca =
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Simple Stresses and
Strains6 o
18 10 m/m/ Cc =
and 6 o12 10 m/m/ Cs =
(b) If the bolt in the assembly showing in Figure 1.26 is loosened by 0.02 mm,calculate the rise in temperature required to induce a stress of 12 N/mm
2in
steel bolt and also calculate the corresponding stress in the copper tube.
(c) For an assembly similar to the one shown in Figure 1.26, if thermal stressesin steel and copper are to be equal what should be the outer diameter of the
copper tube.
Note
Rise in temperature will cause thermal stresses only if of tube material is greaterthan of bolt material. It of tube material is smaller, the assembly will becomeloose due to rise in temperature, while thermal stresses will be induced due to fall
in temperature.
1.6.3 Yielding of Supports
In our earlier analysis of thermal stresses we have assumed the bars to be placed between
rigid supports. As we have already realized that there are no rigid solids, there are also no
rigid supports. The supports may be much stiffer than the bars, but they too undergo
deformation, to a smaller extent, and in many cases these deformations may be too small
to be considered.
Consider the bar shown in Figure 1.25. Without changing any other data, let us assume
the stiffness of the support, ks to be 80 kN/mm (i.e. the support will yield by 1 mm for a
thrust of 80 kN applied on it).
IfP is the force developed then Eq. (1.17) may be rewritten as follows :
t es
Pd
k + = + . . . (1.21)
(Equilibrium requires that P is same in the support and bar.)
Substituting numerical data in Eq. (1.21).
7500.288 0.121200 200 80
P P =
+
(0.0125 0.003125) 0.288 0.12P + =
0.168
10.752 kN0.015625
P = =
Note that in the case of steel bolt and copper tube assembly (Figure 1.26), the steel bolt isserving as any yielding support to the copper tube.
The support has yielded considerably so that most of the stress in the bar is relieved.
SAQ 7(a) Solve the above example with values ofks equal to 200 kN/mm,
500 kN/mm, 1000 kN/mm and 5000 kN/mm. Draw a graph, P vs ks and find
the values ofks for which percentage relief of thrust in the bar are 40 to 25respectively.
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(b) Repeat the above exercise with d= 0 and all other data remaining the same.Stresses in Solids
(c) If the compound bar described in Figure 1.23 is placed just snug betweentwo supports of stiffness 1200 kN/mm and heated up by 20
oC, calculate the
thermal stresses induced.
1.7 STRESSES ON OBLIQUE SECTIONS
We have already seen that the equilibrium of a rigid body should be satisfied overall andin addition if we divide it into a number of small rigid bodies each of these small
elemental bodies should also be in equilibrium individually. The satisfaction ofequilibrium does not depend on the way elements are divided. Even if the solid is divided
by inclined or even curved surfaces, equilibrium must be satisfied for each of theelements thus divided.
Consider the solid shown in Figure 1.27 where the solid is divided into small elements by
inclined planes. The inclination (or orientation) of a plane is defined by its aspect angle,
defined as the angle made by its normal to the longitudinal axis of the original bar. Let
the aspect angle of the plane be . Since the width of the plane b is unaltered and length
of the plane h is increased tocos
h, the area of the inclined plane is
cos
A, whereA is
area of cross section of the original solid.
Ifx be the stress acting on the plane normal tox axis (longitudinal axis), then the axialforce P = xA. This force acting on the inclined plane may be resolved into normal andtangential components.
Normal component on the plane = P cos
= xA cos
Tangential component = P sin
= xA sin
Normal stress on the plane =
cos
cos
A
Ax = x cos2 . . . (1.22)
Shear stress on the plane =
cos
sin
A
Ax = x cos sin . . . (1.23)
a
A
b b d d f
eecca
45O
B C
f h
g
D
P P
O
PP P
PP
PP
45O 60
O
60
Pn2
= 90PoP+
c ePn1
PQtB2BX
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Simple Stresses and
Strains
Figure 1.27
If on the cross sectional area of the original solid shear stress xy is applied, itscomponents on the inclined plane may be evaluated as,
Normal stress component = xy cos sin . . . (1.24)
Shear stress component = xy cos2 . . . (1.25)
By a similar analysis (your exercise), you may verify the following :
If a normal stress ofyis applied on the solid, then the stress components on a plane
whose normal is inclined at to thex axis are given by
Normal stress = y sin2 . . . (1.26)
Shear stress = y sin cos . . . (1.27)
If a shear stress ofyx is applied on the solid, the stress components on the inclined planeare given by (with yx = xy).
Normal stress = xysin cos . . . (1.28)
Shear stress = xysin2 . . . (1.29)
Since a general state of stress in two dimensions is defined by the stress components x,y and xy,as shown in Figure 1.28, general expressions for normal and shear stresscomponents may be obtained by algebraic sum of the respective components fromEqs. (1.22) to (1.29), as given below :
n = x cos2 + y sin
2 +2 xy cos sin
nt= xcos sin + y cos sin + xy(cos2 sin2 )
On further simplification, we obtain
n =22
yxyx ++
cos 2 + xy sin 2 . . . (1.30)
nt=
2
xysin 2 + xy cos 2 . . . (1.31)
y
yx
yx
x
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Stresses in Solids
Figure 1.28 : General State of Stress in Two Dimensions
Application of Eqs.(1.30) and (1.31) will be elaborately dealt with in this unit. However,the significance of the equations should be stressed at this stage. When we are carryingout stress analysis on solids, we may be evaluating stress components on a set of
mutually perpendicular planes, such as x, yand xy. The magnitudes of these
components may not always be sufficient to decide whether the solid is safe or not, andwe require evaluation of stress components on other specific planes, or determination ofplanes on which the stress components have extreme value. Such an analysis can beeasily carried out with the help of Eqs. (1.30) and (1.31).
1.8 RELATIONSHIP BETWEEN ELASTIC
CONSTANTS
In Section 1.3.3, we have defined the four Elastic ConstantsE, , Kand G and also statedthat they are not independent. Now, we shall establish the relationship between them.
1.8.1 Relationship between E and K
Figure 1.29 shows the spherical state of stress in which the normal stress component isthe same, 0 in any direction and shear stress components are zero on any plane. Such a
state of stress is also called as volumetric stress. If the Youngs Modulus and PoissonsRatio of the solid are known asEand , the strain components are given by
x =E
x E
y
E
z . . . (1.32)
Since, x= y = z = 0
x=E
0 (1 2) . . . (1.33)
Similarly, y=E
0 (1 2) and z=E
0 (1 2)
From Eq. (1.6), we can express the change in volume of the solid as
dV= V(x + y + z )
Volumetric strain, v =V
dV= x + y + z
Bulk Modulus, K=strainVolumetricstressVolumetric
=zyx ++
0 =x
3
0
=
)21(30
0
E
K=
)21(3
E. . . (1.34)
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Simple Stresses and
Strains1.8.2 Relationship between E and G
Figure 1.29 shows what is known as the state of pure shear. The deformation of the bodydue to the shear is shown in Figure 1.29 (projected view).
B
G
CD
A
F
O
H
2
+4 2
Figure 1.29
The state of stress on the element is given by the stress component S, xy yx = = .
Due to shearing strain the original planeABCD (a square) gets distorted into a rhombus
EFGH. The magnitude of the shear strain is given by change in the right anglesBAB, ABC, BCD and CDA. This change is produced by rotations of pairs of
lines by2 such as DEH.
Using Eqs. (1.31) and (1.32), let us find the stress components on the diametral planes
DB andACwhose aspect angles are3
and4 4
respectively. On planeDB, we get
Normal stress component, = sin 2
= sin4
2=
Shear stress component = cos 2
= cos4
2= 0
Similarly, on planeAC,
Normal stress component = sin 2
2cos 0
4
= =
Shear stress component = cos 2
3cos 2 0
4= =
Hence, the distortion of the squareABCD by shear stress may also be treated as due to
the elongation of the diagonalACdue to normal stress of + on planes parallel toBDand shortening of the diagonal due to normal stress of on planes parallel toAC.
E
4 2
2
2
2
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Stresses in Solids (1 )OA
vv
E E E
+ = =
. . . (1.35)
(1 )OB
vv
E E E
+ = =
. . . (1.36)
(1 )OAOE OA= +
(1 )OBOF OB= +
4 2OEF
= =
12tan tan
4 21
2
= = +
. . . (1.37)
From Figure 1.29,
(1 ) (1 )tan ( )
(1 ) (1 )
OB OB
OA OA
OB OBOFOA OB
OE OA OB
+ + = = = =
+ +
Q
(1 )1
tan(1 )
1
v
Ev
E
+
= +
+. . . (1.38)
On comparing Eqs. (1.37) and (1.38),
(1 )2
vE
= +
or shear strain2 (1 )v
E
+ =
Rigidity Modulus G =Shear stress
2Shear strain(1 )v
E
=
+
2 (1 )
EG
v=
+. . . (1.39)
1.8.3 Significance of the Relationships
We have already mentioned that there are only two independent elastic constants. But
this is true only in the case of isotropic solids. The term isotropy means same property in
all directions. We have defined Elastic Modulus as ratio between stress and strain and
implied (without stating) that the ratio holds good for all directions, i.e.z
z
y
y
x
x
=
=
.
Materials for which this is true are called isotropic materials and throughout this courseyou will be learning only about isotropic solids.
When any two of the elastic constants are known, the other two may be calculated usingEqs. (1.38) and (1.39). Let us have an example.
Example 1.3
In separate experiments, Youngs Modulus and Rigidity Modulus of a materialhave been determined as 120 GPa and 50 GPa respectively. Calculate the
Poissons Ratio and Bulk Modulus of the material.
Solution
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Simple Stresses and
StrainsHere, Youngs Modulus,E= 120 GPa.
Let Poissons ratio be v.
Rigidity Modulus,2 (1 )
EG
v=
+= 50 GPa
(1 + v) = 502
120
2 =GE
= 1.2
Poissons Ratio, v = 1.2 1 = 0.2
Bulk Modulus, K=3 (1 2 )
E
=)2.021(3
120
= 66.667 GPa.
SAQ 8
(a) Through separate experiments the elastic constants of a material, namely,E,v, G and Kwere determined as 75 GPa, 0.247, 30 GPa and 49.2 GParespectively. Taking the values ofEand G as correct, find the percentage
errors in the values of Poissons Ratio and Bulk Modulus.
(b) The values of Youngs Modulus and Rigidity Modulus of a material areknown to be 20.8 GPa and 8 GPa respectively. If a spherical ball of diameter
150 mm made of the material is immersed in water to a depth of 120 meters,find the change in volume of the ball.
(c) By eliminating Poissons ratio from Eqs. (1.38) and (1.39), show that
G =EK
KE
93
.
1.9 SUMMARY
In this unit you have learnt the concepts of stress, strain, elastic modulus and the basicconcepts in application to analysis of stresses and deformation in simple solids. Any
designer of engineering systems has to have a clear perception of how forces are resistedby solids and the effects produced in solids so as to enable him to produce designs with
safety and stability. The study of these aspects will require learning further new conceptsand methods, understanding of which will require the knowledge you acquired in this
unit. This unit is, therefore, like the first step in a full flight of stairs climbing which willprovide you with a powerful tool of engineering.
1.10 ANSWERS TO SAQs
SAQ 1
(a) Change in volume = 1800 mm3
(b) Youngs ModulusE= 12.7324 kN/mm2
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Poissons Ratio, v = 0.16667.Stresses in Solids
SAQ 2
(a) dv1 = 1462.5 mm3, dv2 = 1950 mm
3, dv3 = 975 mm
3, dv4 = 2437.5 mm
3
Total change in volume dv = 6825 mm3.
(b) (i) For bar given in Figure 1.20, = 24.92 mm
(ii) For bar given in Figure 1.21, = 0.0268 mm
SAQ 3
(b) Equivalent area of copper bar = 13953.91 mm2
SAQ 4
(a) Stress in steel = 7.16645 N/mm2
Stress in copper = 4.3 N/mm2
Stress in aluminium = 2.8666 N/mm2
(b) Load carrying capacity of bar = 837.23 kN.
(c) Area of reinforcement steel required = 3529.4 mm2
(d) The chart will be straight line with the equation
P = 640 1 17 SA
A
+
kN
(e) Load carrying capacity of given column = 1305 kN.
Area of steel required = 5106.4 mm2ifP = 1200 kN.
SAQ 5(a) max = 125.1 N/mm
2, in the 60 mm dia segment.
(b) 1 = 20.644 N/mm2, 2 = 82.577 N/mm
2
3 = 27.526 N/mm2, 4 = 110.103 N/mm
2.
(c) Tmax = 22.88766oC.
SAQ 6
(a) Stress in Aluminium = 24.16 N/mm2
Stress in Copper = 21.84 N/mm2
Stress in Steel = 7.6 N/mm2
(b) Required rise in temperature, T= 43.333oC, andcorresponding stress in copper tube = 9 N/mm
2
(c) Outer diameter of copper tube = 25.456 mm.
SAQ 8
(a) Error in Poissons Ratio = 1.2%
Error in Bulk Modulus = 1.6%
(b) Change in volume of Spherical ball = 122341 mm3or 6.823%.
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Simple Stresses and
Strains
STRESSES IN SOLIDS
This is the first block of three blocks series on Strength of Materials. Block 1 includes
three units and covers certain fundamental principles of strength of materials. This isfollowed by another two blocks by a set of engineering applications of these principles.
Unit 1 introduces basic concepts of simple stress and strain in tensile, compression and
shear. Mechanical behaviour of materials, deformation of bars and composite bars andalso stresses in oblique sections and finally concludes the unit with relations between
different elastic constants.
Unit 2, entitled Thermal Stresses, begins with effects of temperature on bodies bydiscuss their stresses in uniform, stepped and tapered bars. It also analyses thermalstresses in composite/compound bars and tubes.
Unit 3 deals with principal stresses and strains of planes with the identity of plane of
maximum shear stress. It also discusses the Mohrs circle for analysis and the state ofstresses in combined bending and shear, combined bending and torsion. Finally, this unitdiscusses strain energy due to volume and shear stress and also various theories of
failure, concepts of failure and equivalent stresses.
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Stresses in Solids
STRENGTH OF MATERIALS
The mechanics of deformable solids or strength of materials as it is commonly called isone of the core subjects that needs to be studied by all engineering students. In any
manufacturing process, while selecting a suitable material, you are always interested to
know its strength. The strength of a material may be defined as an ability to resist itsfailure and behaviour under the action of external forces. A detailed study of forces and
their effects, along with some suitable protective measures for the safe workingconditions forms the subject matter of Strength of Materials. As a matter of fact, such
knowledge is very essential to enable you to analyse and design all types of structuresand machines. Therefore, the knowledge and practice of Strength of Materials forms an
integral part of learning in all branches of engineering.
This course divided into three blocks and comprises 13 units.
Units 1 and 3, entitled Simple Stresses and Strains and Principal Stresses and Strains
respectively, deal with simple and principal stresses produced in the material on the basisof which an appropriate section of the member (of the given material) is selected for the
construction of given structure. Unit 2 deals with thermal stress in bars and tubes, etc.
Unit 4, entitled Shear Forces and Bending Moments, deals with the most important
parameters used in designing, namely shear force and bending moment. This unit startswith the classification of supports, beams and loadings. It covers all types of possible
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Simple Stresses and
Strainscombinations of loads and beams for shear force and bending moment computationthrough a large number of examples.
Unit 5, entitled Stresses in Beams, covers the details of bending stresses developedacross the cross section of a beam under loading. It also discusses the theory of simple or
pure bending and concept of moment of resistance along with some practical applicationsbased on these concepts. The unit concludes with the concepts of shear stress distribution
in beams.
Unit 6 deals with stress distribution in beams. It discusses section modulus, analysis ofpartial beam section, flitched beams, beams of uniform strength and concludes with shearstress distribution in various beams.
Unit 7, entitled Direct and Bending Stresses, discusses axial load and bending
moments, eccentrically loaded sections and conditions for no tension section.
Unit 8 deals with concepts of strain energy, potential energy, strain energy with suddenlyapplied load. It introduces Castaglianos Theorem and its applications.
Unit 9, entitled Deflection of Beams, presents the concepts of deflection in beamsunder the action of external forces. The unit covers the slope and deflection computations
for cantilever and simply supported beams under standard loadings.
Unit 10, entitled Torsion, discusses deformations undergone by solid circular shaftswhile transmitting power. It also discusses the stresses induced in the circular shafts.
Unit 11 deals with thick and thin cylinders, compound cylinders, and also wire boundpipes, etc.
Unit 12, entitled Shells, deals with stress and strain developed in the thin spherical
shells, doubly curved shells and conical shells.
Unit 13, entitled Springs, deals with closed coiled helical springs, compound springs
and leaf springs with their applications and also introduces study of stress developed inthe springs due to axial load and axial couple.
In each unit, you will find a number of SAQs (Self Assessment Questions) for checking
your own progress. You are advised to study the text carefully, and then attempt SAQson your own and verify your answers with those given at the end of the unit. This willdevelop your confidence in analysing and solving the practical problems.
At the end, we wish you all the best for your all educational endeavours.